In his third week of working here, Tom finally got to sit at his own desk. This week we trained Tom on all of the simulation and analysis software we use here. We went over our swing simulation, aerodynamics, and finite element analysis packages. Finite element analysis was explained by Gilly in a previous blog, but it basically allows us to run performance, stress, and sound analysis on clubs before we make them so we know exactly what to expect when we do get our first prototypes in.

Plus this helps us iterate on different shapes, thicknesses, materials, weights, and all the other club properties before we make any physical parts. In doing so, our initial prototypes are very good and close to our performance targets, then we just have to fine tune them before production. This saves magnitudes of time compared to the old method (prototype a club, test it, modify it, prototype another club, test, modify, repeat until perfect). In learning all of these analysis techniques, Tom will be able to compare his current set of clubs to Cleveland clubs and determine his ideal set (he’ll then have to present his findings to the rest of R&D and defend his statements… his technical interview will seem like a piece of cake compared to this).

I’ve talked mostly about what Tom is doing, but I also want to mention how he’d doing. Tom is picking up this vast amount of new information very well. He is starting to connect the dots between all of the different club properties and variables to see how everything is connected and affects club performance. Here are Tom’s comments on his first week at his own desk.

*As Brian mentioned, I have been sitting at my desk for the last week. It’s nice. My workspace passed its ergonomic assessment so I have no complaints about being there. This past week my “real” training began. It consisted of learning FEA analysis that required me to learn from videos narrated by Brian where he walked me through the entire process, and the following day I learned directly from Brian where he walked me through the entire process. It was a little more in-depth and we began to discuss the results and determine how changes to certain attributes on the club affect the performance of the club.*

*In addition to all of the FEA work that I did, Brian and Alex (another engineer) introduced me to all of the simulation software that is used. In golf rarely do you ever have to hit the same shot twice and consequently we are able to drastically vary the performance characteristics of the clubs in order to mimic the variance in a typical golf outing. I’m currently in the process of using all of this data to determine the best set of Cleveland Golf clubs that would be optimal for me. I feel at times that I’m a statistician at a baseball game trying to come up with some ridiculous information (you all know what I’m talking about). It’s as if I am trying to figure out the best club for me to be hitting on every other Tuesday, after eating a low calorie lunch, with the wind from the Southwest, while wearing pleated pants and sunglasses…or so it seems. We’re working on being able to simulate all of that info. All kidding aside, we are able to generate mass amounts of data using a variety of simulation software programs. I am trying to get as much information as possible to prepare myself for my report presentation which will be next Friday. I really have to determine what is essential to my report and what can be left aside due to time constraints.*

*I’ll let you all know how analyzing this data helps me write my report and prepare myself for going in front of the firing squad. Tune in next week. Good luck with the logic problem…I couldn’t figure it out during my interview.*

You guys are pretty impressive at answering the logic problems, so I’m going to step it up this week with a little tougher one. Congrats to Kyle D. for answering last week’s problem correctly and winning a Cleveland Golf Tour hat. A couple people had the right idea before Kyle, but they didn’t read the question correctly and thought there were multiple lights in the attic when there was actually only one. Also, all those who submitted an answer got entered into the drawing for the “First Run” CG15 Wedge (one of the first 250 CG15 wedges ever made). There are only few more weeks of blogs until the drawing. Below is this week’s logic problem.

*Organization – There’s a group of people in a room, and each one has either a red or blue hat on. No one knows or can see what hat he or she has on, but they can see the color of each other’s hats. One at a time, they have to walk through the door and organize themselves outside by the color of their hat (red hats standing next to each other and blue hats standing next to each other). Without communicating with each other (verbally, non-verbally, or signaling in any way), what process should they use to insure they get organized in their colors. (John – there might be several correct answers so we will pick the first that convinces us it will work)*

(John – quick point -They group can make a strategy before they start the process but once the process starts they can’t communicate. )

All people take their hats off and look at the color, then join the people with their same color hats.

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Have the first person walk out and go to the right. The next person will follow to the right. The third person will go to the right if the people ahead of him match (e.g. both red or both blue) but if they don’t match, he/she will go to the left. Anytime someone goes to the opposite direction, the person prior will change sides.

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Tarek – that is a good method, but how would they communicate this idea of going to left or right when they can’t communicate? The total lack of communication isn’t killing me here.

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IS* killing me. sorry.

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Nice! I did simulation and analysis for a few years a few companies (Raytheon Missle System, White Sands, NASA, Boeing, Northrup)… Good stuff! Keep it coming.

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Let me clarify here. You want to come up with the general strategy or algorithm these people should use, and you would explain it to them beforehand so everyone is on the same page. You just can’t say any specifics about the color of these people’s hats or how many of each color there are, you just want to tell them the process to use.

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Tarek:

Close but no cigar. The last person will have no one behind him to switch him to the correct side.

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I don’t believe there is anyway to organize themselves by color with 100% certainty without any form of communication. However, if everyone were to count all the hats in the room everyone wearing a red hat would see Red-1 red hats and all of the Blue hats. Everyone wearing a blue hat would see Blue -1 and all of the red hats. If everyone (again, unlikely without communication) in the room counted the hats and the hats were divided unequally, the people wearing the hat with the higher number could all leave first and those with the lower number second. However, mathematically, this problem is usually solved with minimal communication.

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Let me further clarify: If after counting everyone’s hats, and Red is greater than Blue, everyone wearing Red hats would leave the room at the same time (lets call this number 10(R-1) seconds. Everyone else, knows they are wearing blue hats and then leaves. If Blue is the greater number, everyone wearing blue leaves at the same time (10 (B-1) seconds).

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in then left right theory the first person would have to look at the last person to finish the segregation of colors

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Use a mirror.

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Strategy Beforehand – since according to the rules they can communicate before the process begins, a captain will be selected for each color.

Process – the first person out the door will have a red hat and go to the right. The second person out the door will have a blue hat and go to the left. Each person thereafter will stop once going through the door. The captain with the matching colored hat will approach the person at the doorway, then they both will turn and go back to that captain’s original spot.

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Its easy. Before the people go in the room, have 2 different people count the hats. This is fact will then tell you what color hat you have on. Example, if the first person counts 50 blue and 49 red hats and the second person counts 49 red and 50 blue, that tells them that the first person has a red hat and the second person has a blue. This method can be used no matter how many people are in the room. Another example, if the first person counts 10 blue and 9 red, and the second person counts 11 blue and 8 red. The first person has blue and the second has red. The next step in group prep is to tell everyone with a blue hat to go to one side of the room and the people with the red hat go to the opposite. This should organize the group with no communication in the second room.

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MY first example was wrong… I meant to say the second person counts 49 blue and 50 red. TYPO

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have 2 pairs of red hats walk out togther and go left then the remaining ppl will go right when they walk out!

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Let’s clarify – You can make a strategy beforehand. i.e. before anyone gets a hat. Once a hat is put on every person, there can be no more verbal or non-verbal signals. Unfortunately this means Terry’s answer won’t work since we can’t select captains because no one knows what color they will be assigned, and Chris’s idea doesn’t work because even though 2 people count the hats they cannot tell anyone that they counted 11 and 8. Though this isn’t the elegant answer Brian had in mind, I think you are heading toward an answer that will work.

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Each person counts the number of red hats and blue hats they see, and waits for 10N seconds, where N is the lower of the two numbers. After that time, (providing the other people are all doing the same thing), he must be wearing the hat of the color which he is seeing fewer of.

Everyone wearing a blue hat will see B − 1 blue hats and R red hats.

Everyone wearing a red hat will see R − 1 red hats and B blue hats.

If B < R then all blue hats will say 'blue' at the same time (10(B − 1) seconds), and everyone else knows they are wearing red.

If R < B then all red hats will say 'red' at the same time (10(R − 1) seconds), and everyone else knows they are wearing blue.

If B = R then all prisoners will know their colour at the same time. (10(R − 1) seconds, or 10(B − 1) seconds, they are equivalent).

They then walk through the door to their respective side.

Pretty much a modification on the axiom of choice.

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This is attributed to Mike O’Conner isn’t it? And it’s on the merits of the Axiom of Choice versus other methods. It’s harder when they can’t talk after going through a room and without the warden (prisoner puzzle where they’re killed for guessing incorrectly) asking the question, or a times going off every 10 seconds for them to get up and leave.

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Okay, I reread my response and it makes little sense. So I’ll try to elaborate and explain it better, maybe.

Let’s assume that the man placed only 1 red hat on my head and 99 blue hats on everyone else’s head. So at the time when the man said start, I would look around and see 99 blue hats. Since there has to be at least one 1 red hat, I would immediately announce that I had a red hat by walking through the door. Everyone else would then know that they had blue hats and would go the opposite way outside.

Now consider the case where I have a red hat and someone else also has a red hat. I would look and see 1 red hat. If I had a blue hat, that person would immediately announce that they had a red hat by walking through the door. But that doesn’t happen. Instead they wait. Now since neither of us was able to announce right away, we know that there must be 2 red hats. So we both announce that we have red hats by walking through the door.

You can follow this pattern with 3 red hats, etc. but it requires that you have a plan in place for timing. Let’s say that everyone will wait ten seconds for every hat they see less of. So if I see no red hats, I will immediately announce “red” after 0 seconds by walking through the door. Everyone else will follow with “BLUE” right after by walking through the door to the other side. Similarly, if I saw 1 red hat, I would plan to say red after 10 seconds and walking through the door. If no one announced after 0 seconds, I would be correct. Then everyone else would say “BLUE” after that by walking through the opposite way.

It doesn’t matter the number of hats or whether there are more or less of one color as long as we wait a discrete period of time to announce. I chose periods of 10 seconds so that there wouldn’t be a chance of mistakes.

If there are R red hats and B blue hats, then:

Everyone wearing a red hat will see R-1 red hats and B blue hats.

Everyone wearing a blue hat will see B-1 blue hats and R red hats.

There are 3 cases:

If R < B then all red hats will say 'red' at the same timeby walking through the door (10(R-1) seconds), and everyone else knows they are wearing blue.

If B < R then all blue hats will say 'Blue' at the same time (10(B-1) seconds) by walking through the door, and everyone else knows they are wearing red.

If R = B then all persons will know their color at the same time. (10(B-1) seconds, or 10(R-1) seconds, they are equivalent).

Let's take an example. If there were 26 red hats and 24 blue hats, and I was wearing a red hat. Then I would plan to say "red" after 260 seconds by walking through the door. However, the people in blue hats would beat me to the punch saying "blue" after 240 seconds by walking through. Then I would say "red" along with all the rest of the people wearing red hats by walking through to the opposite side.

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before receiving the hats, designate 2 people (doesn’t matter which colors they are) to stand next to each other (it doesn’t necessarily matter, but i’m imagining that the people in line are facing the people that are not yet in line). if they are the same color, the next person will line up on the OUTSIDE of the line. participants will continue to line up on the outside until the first person with a different color gets into line. once that occurs, the people will then line up BETWEEN the groups of opposing colors. after all of the people are in line, the second to last person will know which color they are.

to exit the room, the final member will leave first. the second to the last person that got into line will be the second person to leave the room. they should know whether to group with the 1st exiting person or not because they should have been able to determine their color already. after the 2nd person makes their determination and leaves the room appropriately, the rest can leave the room to their appropriate sides.

here’s an example:

odd numbers are the red group

even numbers are the blue group

[ ] = the formation

{ } = unfiltered group

= person entering the formation

[1 3] {0 2 4 5 6 7 8 9}

[1 3 ] {0 2 4 6 7 8 9}

[1 3 5 ] {2 4 6 7 8 9}

[1 3 5 0] {4 6 7 8 9}

[1 3 5 2 0] {6 7 8 9}

[1 3 5 4 2 0] {7 8 9}

[1 3 5 6 4 2 0] {8 9}

[1 3 5 7 6 4 2 0] {9}

[1 3 5 7 8 6 4 2 0] { } all people are lined up now

9 exits the room first.

8 knows that he is blue because 9 got between him and 7, and 8 is standing next to 6 (which is blue).

7 knows that he is red because he was standing between 5 and 9 (both red).

the rest of the people can leave appropriately. this should work even if there aren’t an equal number of red and blue hats.

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some of my spacing got truncated, so the example may be a bit more confusing.

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–Oh Man!, Horrible question –

Note: The group can make a strategy before they start the process but once the process starts they can’t communicate:

Attempt:

-personA tells all the people with a blue hat on to cross their arms and walk outside

-personB tells personA to cross his arms if he is wearing a blue hat and walk outside to join the rest of the people with crossed arms

-If personA does not have to cross his arms than he will join the group with no crossed arms and walk outside.

If they follow this strategy than you will have all of the people outside with crossed arms together and should be wearing a blue hat. All of the people who walk out with their arms down will join together and should all be wearing a red hat.

This method does not involve telling anyone which color hat they ahve on (directly) and does not involve any communication after the initial plan.

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Wow, John keeps changing or adding rules as it goes along. I sure wish I was smart enough to answer this or any of the other preceeding questions. But I would like to win something.

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Since the original question was not specific about the hats, most hats are the same color on the top of the bil as on the bottom. So all you would need to do is glance up and see what color your hat is and follow that person to the side they represent. I know this is not the exact answer you are looking for but with the information givin this would be a very simple conclusion!!!

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1) Each one is to decide whether he/she is wearing a blue or red hat.

2) If one thinks he wears a red hat, he turns right when he walks out of the door. For a blue hat, he turns left.

3) Once on the picked side, he would form a line with whoever else was already on that side, shoulder to shoulder, facing the door.

3) Once the person goes to his side, he then looks at all the other people on that side (if any). If he finds a person with the wrong color hat for that side, he stands behind that person. The person goes back thru the door to the starting point, and the new person takes his spot. When this person’s turn comes up again, he of course goes to the other side.

4) This goes on until all people have filed to their proper sides.

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For sake of argument let’s say there are 10 people in this organization. The strategy they will communicate is this:

If you are sure you have a red hat then leave the room, if you are sure you have a blue hat or you aren’t sure of your hat color then you stay in the room. The catch is you can only leave every minute on the minute.

For example let’s say there are 9 blue hats and 1 red hat. The person with the red hat will look around and see no red hats so they will realize they are the only one with a red hat and leave at the 1st minute, and everyone else will realize that they have a blue hat.

Now there are 8 blue hats and 2 red hats, the two people with the red hats will look around and see 1 red hat and 8 blue hats, and think to themselves I cannot determine what color my hat is, but if the person with the 1 red hat see’s all blue hats they will leave at the 1st minute and I will know that I have a blue hat. If they do not leave at the first minute they aren’t the only red hat, so I must have a red hat, and both people will get up and leave.

Now say there are 5 red hats and 5 blue hats. The 5 people with red hats will look around and see 5 blue hats and 4 red hats. They will think, I have either a red hat or a blue hat. If this particular person has a blue hat then the 4 red hats will all leave in the 4th minute. But if this person has a red hat, no one will leave at the 4th minute, because everyone else can see 4 red hats, in which case the person must conclude they have a red hat. So the 5 people with red hats will all stand up at 5th minute and leave.

So each individual will count the number of red hats they see and declare that number X, if no one has left by the Xth minute, then they have a red hat. If everyone with a red hat, that they can see leaves on the Xth minute then they have a blue hat. Or in other words you must leave on the X+1 minute if no one has left yet.

So in general, if there are M blue hats and N red hats. At the Nth minute everyone with red hats will stand up because they know that there must be N red hats. And at that point they will all be sorted into red and blue hats.

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Have a strategy and select one person who starts sorting process. Each prson has s number assigned to them (2,3,4, etc). Person one walks over to a person who has a red hat, person two does same, then three, etc. Until thre are no red hats left, all others have blue and exit! Done.

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Everyone just takes their hats off and then looks at the color, then organize into blue or red groups

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They should just do the hustle!

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as a group establish two positions “the enforcer and “the closer”. “the enforcer” would line every one up, once he/she had everyone lined up “the closer” would step out of said line and place “the enforcer” in the proper position.

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One person organize the people into red and blue groups. Then at the end, the people see from the other people in the group what color they are then they show the organizer what group he is in. simple.

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Strategy session – Everybody gets a number 1 thru what ever, 1 and 2 would go to the door and and stand side by side, number 3 would only move towards the door if they had different colored hats. If 3 didn’t move, then 1 and 2 would know they had the same colored hat and then would stand in line. Once they stood in line 3 would move to stand aside 1, 4 would look and if this was correct he would go and stand in line with 1 and 2 or 3 (doesn’t matter) but 5 wouldn’t move until 4 would be correct line, this would continue till the last person stood in line. At that time everybody would know they were standing in the red or blue line except the last person. #1 would not proceed thru the door until the last person was in the correct line.

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The strategy is that everyone finds a partner. Then each pair migrate toward a pair whose hats match each other. If no-one joins your pair, you have a different hat to the person you chose. So if you chose blue, you have a red hat, go to the group with red hats. If someone joined your group you have the same colour as your partner. Now you know your colour, form two groups based on what colour hat you have.

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I suggest a rowdy square dance..

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